Divide the following rational expressions and simplify the result. $\dfrac{9-121k^2}{11k^2+3k} \div \dfrac{11k^2+19k-6}{8k^4+8k^3-16k^2}=$
Explanation: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $9-121k^2$, of the first expression can be factored as $(3+11k)(3-11k)$ using the difference of squares pattern. The denominator, $11k^2+3k$, of the first expression can be factored as $k(11k+3)$ by factoring out a $k$. The numerator, $11k^2+19k-6$, of the second expression can be factored by grouping as $(11k-3)(k+2)$. The denominator, $8k^4+8k^3-16k^2$, of the second expression can be factored as $8k^2(k+2)(k-1)$ by factoring out $8k^2$ and using the sum-product pattern. Now the quotient looks as follows: $\dfrac{(3+11k)(3-11k)}{k(11k+3)} \div \dfrac{(11k-3)(k+2)}{8k^2(k+2)(k-1)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(3+11k)(3-11k)}{k(11k+3)} \div \dfrac{(11k-3)(k+2)}{8k^2(k+2)(k-1)}$ $\begin{aligned}&= \dfrac{(3\!+\!11k)(3\!-\!11k)}{k(11k+3)} \!\cdot\! \dfrac{8k^2(k\!+\!2)(k\!-\!1)}{(11k\!-\!3)(k\!+\!2)} &\text{Flip the divisor.}\\\\\\ &= \dfrac{(3\!+\!11k)(3\!-\!11k) \cdot 8k^2(k\!+\!2)(k\!-\!1)}{k(11k+3) \cdot (11k-3)(k+2)} &\text{Multiply across.}\\\\\\ &= \dfrac{\!\!\!{\cancel{(3\!+\!11k)}}\!\!\!\!{\cancel{(3\!-\!11k)}} \!\!8k\!{\cancel{(k)}}\!\!\!{\cancel{(k\!+\!2)}}\!\!(k\!-\!1)}{{\cancel{k}}{\cancel{(11k+3)}}\!\!(-1)\!\!{\cancel{ (3-11k)}}\!\!\!{\cancel{(k+2)}}} &\text{Cancel out common factors.}\\\\\\\\ &=-8k(k-1) \end{aligned}$ Therefore, the simplified form of the quotient is $-8k(k-1)$, which is equivalent to $8k-8k^2$.